Flipping The Unusual Coins

You have three coins. One always comes up heads, one always comes up tails, and one is just a regular coin (has equal change of heads or tails). If you pick one of the coins randomly and flip it twice and get heads twice, what is the chance of flipping heads again?

Chances of flipping head again are - .......% Click to know!

Chance of Flipping Head Again

What was the problem?

For a coin to always show head on flip we assume both it's sides are heads and the coin which is showing tail always we assume both of it's sides are tails.

There is no way that you have selected tail only coin since there are 2 heads in first 2 flips.

So it could be either head only coin say D coin or regular fair coin say F.

Let H1 and H2 be the sides of head coin and H, T are side of fair coin.

If it's head only coin D, then possible scenarios on 2 flips are -

DH1 DH1
DH1 DH2
DH2 DH1
DH2 DH2

And if it's fair coin F then possible scenarios on 2 flips are -

FH FH
FH FT
FT FH
FT FT

There are total five combinations (all 4 of head coin + first one of fair coin) where there are 2 consecutive heads on 2 flips.

So, the chances that you have picked a head coin is (4/5) and that you picked fair coin is (1/5).

For head coin, the probability of getting head again is 1 and that for fair coin is (1/2).

Since you holding either head coin or fair coin,

Probability (Head on third flip) = 
Probability (You picked Head coin) x Probability (Head on head coin) + Probability (You picked fair coin) x Probability (Head on fair coin) 


Probability (Head on third flip) = (4/5) x 1 + (1/5) x (1/2)

Probability (Head on third flip) = 9/10.

Hence, the chance of flipping head again on third flip is 90%.

Unfair Game of Strange dice

Katherine and Zyan are playing a game using strange dice. Each die is a cube with six sides. Katherine's die has sides numbered 3, 3, 3, 3, 3, and 6. Zyan's die has sides numbered 2, 2, 2, 5, 5, and 5.

To play the game, Katherine and Zyan roll their dice at the same time and whoever rolls the higher value wins. If they play many times, who will win more frequently, Katherine or Zyan?

This person will be winning more! 
 

Advantage in Unfair Game of Strange Dice

What was the game?

Shortest Way : 

Katherine's die has sides numbered 3, 3, 3, 3, 3, and 6. And Zyan's die has sides numbered 2, 2, 2, 5, 5, and 5. That means whenever Zyan rolls a 2 then Katherine will always win. The probability that Zyan rolls 2 is 3/6 = 1/2. So in half of the cases, the Katherine will be winner of the game. Moreover, whenever, Zyan rolls a 5, Katherine can win if she rolls a 6. In short, in more than half cases, Katherine will win hence she has more advantage in this game.


Precise Way : 

Katherine will win if

1. Zyan rolls a 2 with probability 3/6 = 1/2

2. Zyan rolls a 5 (probability 3/6 = 1/2) and Katherine rolls a 6 (probability 1/6). So the probability for this win = 1/12.

Hence, Katherine has got 1/2 + 1/12 = 7/12 chances of winning.

Zyan will win if he rolls a 5 (probability 1/2) and Katherine rolls a 3 (probability 5/6) with probability = 1/2 x 5/6 = 5/12.  

That proves, Katherine has more chances (7/12 vs 5/12) of winning this game. 

Another Way : 

There can be 36 possible cases of numbers of cubes out of which in 15 cases Zyan seems to be winning and in 21 cases, Katherine is winning. Again, Katherine having more chances of winning (21/36 = 7/12) than Zyan (15/36 = 5/12). 

The Tuesday Birthday Problem

I ask people at random if they have two children and also if one is a boy born on a Tuesday. After a long search I finally find someone who answers yes. What is the probability that this person has two boys? Assume an equal chance of giving birth to either sex and an equal chance to giving birth on any day.

Tip: Don't conclude too early. 

Click here to know the correct answer! 

Finding The Correct Probability

How tricky it was?

If you think that the probability is 1/2 after reading that the couple has equal chance of having child of either sex then you are in wrong direction.

Take a look at the table below.

There are 27 possible combinations when boy is born on Tuesday. Out of which there are only 13 possible combinations where either boy (first or second) is born on Tuesday. 

Hence the probability that the person having at least 1 boy off his 2 boys born on Tuesday is 13/27.